### 3.491 $$\int e^{\coth ^{-1}(a x)} \sqrt{c-\frac{c}{a x}} x \, dx$$

Optimal. Leaf size=124 $\frac{c x^2 \sqrt{1-\frac{1}{a^2 x^2}}}{2 \sqrt{c-\frac{c}{a x}}}+\frac{c x \sqrt{1-\frac{1}{a^2 x^2}}}{4 a \sqrt{c-\frac{c}{a x}}}-\frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-\frac{1}{a^2 x^2}}}{\sqrt{c-\frac{c}{a x}}}\right )}{4 a^2}$

[Out]

(c*Sqrt[1 - 1/(a^2*x^2)]*x)/(4*a*Sqrt[c - c/(a*x)]) + (c*Sqrt[1 - 1/(a^2*x^2)]*x^2)/(2*Sqrt[c - c/(a*x)]) - (S
qrt[c]*ArcTanh[(Sqrt[c]*Sqrt[1 - 1/(a^2*x^2)])/Sqrt[c - c/(a*x)]])/(4*a^2)

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Rubi [A]  time = 0.239776, antiderivative size = 124, normalized size of antiderivative = 1., number of steps used = 5, number of rules used = 5, integrand size = 23, $$\frac{\text{number of rules}}{\text{integrand size}}$$ = 0.217, Rules used = {6178, 863, 873, 875, 208} $\frac{c x^2 \sqrt{1-\frac{1}{a^2 x^2}}}{2 \sqrt{c-\frac{c}{a x}}}+\frac{c x \sqrt{1-\frac{1}{a^2 x^2}}}{4 a \sqrt{c-\frac{c}{a x}}}-\frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-\frac{1}{a^2 x^2}}}{\sqrt{c-\frac{c}{a x}}}\right )}{4 a^2}$

Antiderivative was successfully veriﬁed.

[In]

Int[E^ArcCoth[a*x]*Sqrt[c - c/(a*x)]*x,x]

[Out]

(c*Sqrt[1 - 1/(a^2*x^2)]*x)/(4*a*Sqrt[c - c/(a*x)]) + (c*Sqrt[1 - 1/(a^2*x^2)]*x^2)/(2*Sqrt[c - c/(a*x)]) - (S
qrt[c]*ArcTanh[(Sqrt[c]*Sqrt[1 - 1/(a^2*x^2)])/Sqrt[c - c/(a*x)]])/(4*a^2)

Rule 6178

Int[E^(ArcCoth[(a_.)*(x_)]*(n_.))*((c_) + (d_.)/(x_))^(p_.)*(x_)^(m_.), x_Symbol] :> -Dist[c^n, Subst[Int[((c
+ d*x)^(p - n)*(1 - x^2/a^2)^(n/2))/x^(m + 2), x], x, 1/x], x] /; FreeQ[{a, c, d, p}, x] && EqQ[c + a*d, 0] &&
IntegerQ[(n - 1)/2] && IntegerQ[m] && (IntegerQ[p] || EqQ[p, n/2] || EqQ[p, n/2 + 1] || LtQ[-5, m, -1]) && In
tegerQ[2*p]

Rule 863

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> Simp[((d + e*x
)^m*(f + g*x)^(n + 1)*(a + c*x^2)^p)/(g*(n + 1)), x] + Dist[(c*m)/(e*g*(n + 1)), Int[(d + e*x)^(m + 1)*(f + g*
x)^(n + 1)*(a + c*x^2)^(p - 1), x], x] /; FreeQ[{a, c, d, e, f, g}, x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e
^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0] && GtQ[p, 0] && LtQ[n, -1] &&  !(IntegerQ[n + p] && LeQ[n + p + 2, 0]
)

Rule 873

Int[((d_) + (e_.)*(x_))^(m_)*((f_.) + (g_.)*(x_))^(n_)*((a_) + (c_.)*(x_)^2)^(p_), x_Symbol] :> -Simp[(e^2*(d
+ e*x)^(m - 1)*(f + g*x)^(n + 1)*(a + c*x^2)^(p + 1))/((n + 1)*(c*e*f + c*d*g)), x] - Dist[(e*(m - n - 2))/((n
+ 1)*(e*f + d*g)), Int[(d + e*x)^m*(f + g*x)^(n + 1)*(a + c*x^2)^p, x], x] /; FreeQ[{a, c, d, e, f, g, m, p},
x] && NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0] &&  !IntegerQ[p] && EqQ[m + p, 0] && LtQ[n, -1] && IntegerQ[
2*p]

Rule 875

Int[Sqrt[(d_) + (e_.)*(x_)]/(((f_.) + (g_.)*(x_))*Sqrt[(a_) + (c_.)*(x_)^2]), x_Symbol] :> Dist[2*e^2, Subst[I
nt[1/(c*(e*f + d*g) + e^2*g*x^2), x], x, Sqrt[a + c*x^2]/Sqrt[d + e*x]], x] /; FreeQ[{a, c, d, e, f, g}, x] &&
NeQ[e*f - d*g, 0] && EqQ[c*d^2 + a*e^2, 0]

Rule 208

Int[((a_) + (b_.)*(x_)^2)^(-1), x_Symbol] :> Simp[(Rt[-(a/b), 2]*ArcTanh[x/Rt[-(a/b), 2]])/a, x] /; FreeQ[{a,
b}, x] && NegQ[a/b]

Rubi steps

\begin{align*} \int e^{\coth ^{-1}(a x)} \sqrt{c-\frac{c}{a x}} x \, dx &=-\left (c \operatorname{Subst}\left (\int \frac{\sqrt{1-\frac{x^2}{a^2}}}{x^3 \sqrt{c-\frac{c x}{a}}} \, dx,x,\frac{1}{x}\right )\right )\\ &=\frac{c \sqrt{1-\frac{1}{a^2 x^2}} x^2}{2 \sqrt{c-\frac{c}{a x}}}-\frac{\operatorname{Subst}\left (\int \frac{\sqrt{c-\frac{c x}{a}}}{x^2 \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{4 a}\\ &=\frac{c \sqrt{1-\frac{1}{a^2 x^2}} x}{4 a \sqrt{c-\frac{c}{a x}}}+\frac{c \sqrt{1-\frac{1}{a^2 x^2}} x^2}{2 \sqrt{c-\frac{c}{a x}}}+\frac{\operatorname{Subst}\left (\int \frac{\sqrt{c-\frac{c x}{a}}}{x \sqrt{1-\frac{x^2}{a^2}}} \, dx,x,\frac{1}{x}\right )}{8 a^2}\\ &=\frac{c \sqrt{1-\frac{1}{a^2 x^2}} x}{4 a \sqrt{c-\frac{c}{a x}}}+\frac{c \sqrt{1-\frac{1}{a^2 x^2}} x^2}{2 \sqrt{c-\frac{c}{a x}}}+\frac{c^2 \operatorname{Subst}\left (\int \frac{1}{-\frac{c}{a^2}+\frac{c^2 x^2}{a^2}} \, dx,x,\frac{\sqrt{1-\frac{1}{a^2 x^2}}}{\sqrt{c-\frac{c}{a x}}}\right )}{4 a^4}\\ &=\frac{c \sqrt{1-\frac{1}{a^2 x^2}} x}{4 a \sqrt{c-\frac{c}{a x}}}+\frac{c \sqrt{1-\frac{1}{a^2 x^2}} x^2}{2 \sqrt{c-\frac{c}{a x}}}-\frac{\sqrt{c} \tanh ^{-1}\left (\frac{\sqrt{c} \sqrt{1-\frac{1}{a^2 x^2}}}{\sqrt{c-\frac{c}{a x}}}\right )}{4 a^2}\\ \end{align*}

Mathematica [A]  time = 0.324908, size = 148, normalized size = 1.19 $\frac{2 a^2 x^2 \sqrt{1-\frac{1}{a^2 x^2}} (2 a x+1) \sqrt{c-\frac{c}{a x}}+\sqrt{c} (1-a x) \log \left (2 a^2 \sqrt{c} x^2 \sqrt{1-\frac{1}{a^2 x^2}} \sqrt{c-\frac{c}{a x}}+c \left (2 a^2 x^2-a x-1\right )\right )+\sqrt{c} (a x-1) \log (1-a x)}{8 a^2 (a x-1)}$

Warning: Unable to verify antiderivative.

[In]

Integrate[E^ArcCoth[a*x]*Sqrt[c - c/(a*x)]*x,x]

[Out]

(2*a^2*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)]*x^2*(1 + 2*a*x) + Sqrt[c]*(-1 + a*x)*Log[1 - a*x] + Sqrt[c]*(1
- a*x)*Log[2*a^2*Sqrt[c]*Sqrt[1 - 1/(a^2*x^2)]*Sqrt[c - c/(a*x)]*x^2 + c*(-1 - a*x + 2*a^2*x^2)])/(8*a^2*(-1 +
a*x))

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Maple [A]  time = 0.17, size = 102, normalized size = 0.8 \begin{align*} -{\frac{x}{8}\sqrt{{\frac{c \left ( ax-1 \right ) }{ax}}} \left ( -4\,{a}^{3/2}x\sqrt{ \left ( ax+1 \right ) x}-2\,\sqrt{ \left ( ax+1 \right ) x}\sqrt{a}+\ln \left ({\frac{1}{2} \left ( 2\,\sqrt{ \left ( ax+1 \right ) x}\sqrt{a}+2\,ax+1 \right ){\frac{1}{\sqrt{a}}}} \right ) \right ){\frac{1}{\sqrt{{\frac{ax-1}{ax+1}}}}}{a}^{-{\frac{3}{2}}}{\frac{1}{\sqrt{ \left ( ax+1 \right ) x}}}} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

int(1/((a*x-1)/(a*x+1))^(1/2)*x*(c-c/a/x)^(1/2),x)

[Out]

-1/8/((a*x-1)/(a*x+1))^(1/2)*(c*(a*x-1)/a/x)^(1/2)*x/a^(3/2)*(-4*a^(3/2)*x*((a*x+1)*x)^(1/2)-2*((a*x+1)*x)^(1/
2)*a^(1/2)+ln(1/2*(2*((a*x+1)*x)^(1/2)*a^(1/2)+2*a*x+1)/a^(1/2)))/((a*x+1)*x)^(1/2)

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Maxima [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c - \frac{c}{a x}} x}{\sqrt{\frac{a x - 1}{a x + 1}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)*x*(c-c/a/x)^(1/2),x, algorithm="maxima")

[Out]

integrate(sqrt(c - c/(a*x))*x/sqrt((a*x - 1)/(a*x + 1)), x)

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Fricas [A]  time = 2.20489, size = 674, normalized size = 5.44 \begin{align*} \left [\frac{{\left (a x - 1\right )} \sqrt{c} \log \left (-\frac{8 \, a^{3} c x^{3} - 7 \, a c x - 4 \,{\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + a x\right )} \sqrt{c} \sqrt{\frac{a x - 1}{a x + 1}} \sqrt{\frac{a c x - c}{a x}} - c}{a x - 1}\right ) + 4 \,{\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + a x\right )} \sqrt{\frac{a x - 1}{a x + 1}} \sqrt{\frac{a c x - c}{a x}}}{16 \,{\left (a^{3} x - a^{2}\right )}}, \frac{{\left (a x - 1\right )} \sqrt{-c} \arctan \left (\frac{2 \,{\left (a^{2} x^{2} + a x\right )} \sqrt{-c} \sqrt{\frac{a x - 1}{a x + 1}} \sqrt{\frac{a c x - c}{a x}}}{2 \, a^{2} c x^{2} - a c x - c}\right ) + 2 \,{\left (2 \, a^{3} x^{3} + 3 \, a^{2} x^{2} + a x\right )} \sqrt{\frac{a x - 1}{a x + 1}} \sqrt{\frac{a c x - c}{a x}}}{8 \,{\left (a^{3} x - a^{2}\right )}}\right ] \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)*x*(c-c/a/x)^(1/2),x, algorithm="fricas")

[Out]

[1/16*((a*x - 1)*sqrt(c)*log(-(8*a^3*c*x^3 - 7*a*c*x - 4*(2*a^3*x^3 + 3*a^2*x^2 + a*x)*sqrt(c)*sqrt((a*x - 1)/
(a*x + 1))*sqrt((a*c*x - c)/(a*x)) - c)/(a*x - 1)) + 4*(2*a^3*x^3 + 3*a^2*x^2 + a*x)*sqrt((a*x - 1)/(a*x + 1))
*sqrt((a*c*x - c)/(a*x)))/(a^3*x - a^2), 1/8*((a*x - 1)*sqrt(-c)*arctan(2*(a^2*x^2 + a*x)*sqrt(-c)*sqrt((a*x -
1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x))/(2*a^2*c*x^2 - a*c*x - c)) + 2*(2*a^3*x^3 + 3*a^2*x^2 + a*x)*sqrt((a*x
- 1)/(a*x + 1))*sqrt((a*c*x - c)/(a*x)))/(a^3*x - a^2)]

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Sympy [F(-1)]  time = 0., size = 0, normalized size = 0. \begin{align*} \text{Timed out} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))**(1/2)*x*(c-c/a/x)**(1/2),x)

[Out]

Timed out

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Giac [F]  time = 0., size = 0, normalized size = 0. \begin{align*} \int \frac{\sqrt{c - \frac{c}{a x}} x}{\sqrt{\frac{a x - 1}{a x + 1}}}\,{d x} \end{align*}

Veriﬁcation of antiderivative is not currently implemented for this CAS.

[In]

integrate(1/((a*x-1)/(a*x+1))^(1/2)*x*(c-c/a/x)^(1/2),x, algorithm="giac")

[Out]

integrate(sqrt(c - c/(a*x))*x/sqrt((a*x - 1)/(a*x + 1)), x)